Question

Calculate capacitance​

Answer 1

Answer:

$\\\tt \dfrac{2A \epsilon_{\circ} ( \kappa_{1} \kappa_{1} ) }{\kappa_{1} + \kappa_{1}}$

Explanation:

Since $\\\tt \kappa _{1} , \kappa _{1}$ are in direct connection they are considered to be in series,

Let $\\\tt C_{1}$ have $\\\tt \kappa_{1}$  and $\\\tt C_{2}$ have $\\\tt \kappa_{2}$

Now for series combination of two capacitors is given by:

$\\\tt \dfrac{1}{C_{eq}} =\dfrac{1}{C_{1}} + \dfrac{1}{C_{2}} \\ \\ =\dfrac{C_{1} +C_{2} }{C_{1}C_{2}}$

Now,

$\\\tt C_{eq} =\dfrac{C_{1} C_{2} }{C_{1} +C_{2}}$

Since,

$\\\tt C= \dfrac{2 A \epsilon_{\circ} \kappa }{d}$ here.

Using this in the net capacitance expression:

$\\\tt C_{eq} = \dfrac{C_{1} C_{2} }{C_{1} +C_{2}} \\\\ =\dfrac{\dfrac{ 2A \epsilon_{\circ} \kappa_{1} }{d} \times \dfrac{2A \epsilon_{\circ} \kappa_{2} }{d} }{\dfrac{2A \epsilon_{\circ} \kappa_{1} }{d} + \dfrac{2A \epsilon_{\circ} \kappa_{2} }{d} } \\ \\= \dfrac{2A \epsilon_{\circ} ( \kappa_{1} \kappa_{1} ) }{\kappa_{1} + \kappa_{1}}$